Need help with Mechanic Energy math problem.

[Anhkha]

Homework Statement

"A little girl is sliding down a snowy hill on her sledge and out on a horizontall flat area. The girl has a physics interested mom which wants to know what the maxium speed the sledge can reach. She measures that the hill is 3,0m high and 6,0m long. The sledge slides for 12.5m before stoping. She belives that the friction power on the hill is 85% of hte friction power of the flat area. What is the highest maxium speed the sledge can reach?

The answer is supposed to be 6.5m/s

This is my terible drawing i made

Homework Equations

W=-Rs
E_k =1/2 mv^2
E_p= mgh
E_k+E_p=E_k0+E_p0+W

The Attempt at a Solution

At the end i reach
V=sqrt(mgh/(1.85*0.888)), which equals to the wrong answer..
I don't even know. i have tried to solve this for like 2 hours now and its driving me crazy.

Any help would be apprecieated a lot.

Sorry for the bad english, i translated the problem from Norwegian to English.

 

[TomHart]

So you drew the picture from the wording of the problem? What I questioned is if the flat portion is 12.5 meters or 6.5 meters. Is it clear in the original language?

 

[Anhkha]

So you drew the picture from the wording of the problem? What I questioned is if the flat portion is 12.5 meters or 6.5 meters. Is it clear in the original language?

The flat portion is 12.5m.
6.5m is from the diagonal line from the top of the hill to the flat area. Sorry for late answer, just took bus from library to home.
Maybe this drawing can clear some things up.

 

[TomHart]

Thank you. There is enough information to reach a final numeric answer. So your final answer should be a number only (plus units).

So the basic problem is that you start out with all of the sled's energy being potential energy. The sled accelerates down the hill and all of that potential energy goes to one of two places - either converted to kinetic energy of the sled or is dissipated (Work). So at the bottom of the hill, the sled will only have kinetic energy. At the end of the flat portion, all of that kinetic energy will have been dissipated (Work) and the sled will have v = 0 m/s. So it looks like this is a problem where you have to solve both portions (hill and flat) simultaneously, because there is not enough information given to solve either part by itself.

Can you show some of your work - how to start the problem?

 

[Anhkha]

Here is some of my calulations, warning: this might be completely wrong.
EDIT: s=distance. S is distance in my language :)

 

[Anhkha]

Thank you. There is enough information to reach a final numeric answer. So your final answer should be a number only (plus units).

So the basic problem is that you start out with all of the sled's energy being potential energy. The sled accelerates down the hill and all of that potential energy goes to one of two places - either converted to kinetic energy of the sled or is dissipated (Work). So at the bottom of the hill, the sled will only have kinetic energy. At the end of the flat portion, all of that kinetic energy will have been dissipated (Work) and the sled will have v = 0 m/s. So it looks like this is a problem where you have to solve both portions (hill and flat) simultaneously, because there is not enough information given to solve either part by itself.

Can you show some of your work - how to start the problem?

I posted some of my calculations above which might be completely wrong. However i will try to do make some more calulculatiions i can post.

 

[Anhkha]

Known
h₀=3.0m
d₁=6.0 m (From top of the hill diagonaly down to the flat area)
d₂= 12.5 m (Flat area)
E_p=0J (Since no potenial energy at the end of the ride)
E_k0=0j (Since no movement at the top of the hill)
Friction on d₁ is 85% of the friction power to d₂
Unknown
V(Velocity)

 

[TomHart]

Check your math in the second to the last row of your work. You have 0.888 in the denominator. That doesn't look right according to my calculations.

Edit: Actually, what you did is an illegal math operation going from the previous row to that row. You can't add 1 + 0.85 in that situation.

 

[Anhkha]

isnt 1.85*6

Check your math in the second to the last row of your work. You have 0.888 in the denominator. That doesn't look right according to my calculations.

Edit: Actually, what you did is an illegal math operation going from the previous row to that row. You can't add 1 + 0.85 in that situation.

Isnt 1.85*(6/12.5)=0.888?
1.85*(0.48)=0.888?

 

[Anhkha]

Check your math in the second to the last row of your work. You have 0.888 in the denominator. That doesn't look right according to my calculations.

Edit: Actually, what you did is an illegal math operation going from the previous row to that row. You can't add 1 + 0.85 in that situation.

Oh! Didnt see that!i Will try to see if i get something else. Thank you
EDIT: Solved! Cant believe i used to hours to figure out a stupid mistake like that haha. Thank you a lot!

 

[TomHart]

1 + (0.85)(6/12.5) ≠ (1 + 0.85)(6/12.5)

Edit: Never mind. I see that you got it.